Revision as of 06:48, 17 June 2009 by Vramani (Talk | contribs)

$ x(t)=\sqrt{t} $


$ E\infty= \int_{-\infty}^{\infty}|x(t)|^2dt $

E$ \infty $ = $ \int_{-\infty}^{\infty} tdt $


E$ \infty $ = $ \frac{1}{2} t^2 $ evaluated from -$ \infty $ to +$ \infty $ = $ \infty $


P$ \infty $ = lim T$ \to $$ \infty $ $ \frac{1}{2T} $ $ \int_{-T}^{T}\ tdt $

$ \frac{1}{2T} (.5t^2)|_{-T}^{T} = \frac{T}{4} $

lim T$ \to $$ \infty $ = $ \infty $ = P$ \infty $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang